Question

triangle xyz is shown on the coordinate plane below: triangle xyz on the coordinate plane with ordered pairs at x 3, negative1, at y 4, negative 4, at z 1, negative 2 if triangle xyz is reflected across the line y = 1 to create triangle x'y'z', what is the ordered pair of x'?

Answer 1

Original points are: x = (3, -1); y = (4,-4); z = (1,-2). A reflection across the line y = 1 (a line parallel to the y-axis) will keep the same x-coordinates and will move the y-coordinate upward as many units above the line y = 1 as units are from the line y = 1 to the y-coordinate of the original point. Distance of (3,-1) to the line y = 1 is 1 - (-1) = 2 => new y-coordinate = 1 + 2 = 3. So, the answer is that the ordered pair ox x' is (3,3)

Answer 2

Answer:

x' (3, 3)

Step-by-step explanation:

Reflection across the line y = k, where k is some known constant (here k = 1), transform points (x, y) to (x, -y+2*k). Then points x, y and z create triangle x'y'z' as follows:

x (3, -1) -> x' (3, 3)

y (4, -4) -> y' (4, 6)

z (1, -2) -> z' (1, 4)